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Piecewise Boundaries and Their Breakpoints

Learn to calculate areas with piecewise upper, lower, left, or right boundaries by splitting at every formula breakpoint and verifying boundary order on each interval.

Intermediate 10 min read

What You'll Learn

  • Identify the breakpoints of piecewise-defined boundaries on an area interval
  • Combine the breakpoints from two boundaries into a single partition
  • Set up separate area integrals on intervals where the boundary formulas are fixed
  • Check that upper and lower, or right and left, boundaries are correctly ordered
  • Calculate and verify areas by adding the pieces

Why Piecewise Boundaries Need a Plan

A region can have a boundary that changes its formula partway across the region. On one interval, its upper edge might be a line; on another, it might be a different line or curve. The same issue arises for left and right edges when we use horizontal slices.

The area method is still the same on each interval: integrate the distance between the two boundaries. The extra work is to identify where a formula changes, split the integral there, and make sure the boundary order is correct on every resulting piece.

Key idea: A single integral formula applies only where the boundary formulas and their order are fixed. Partition the interval at every relevant breakpoint, calculate the area on each part, then add those areas.

Breakpoints and Piecewise Boundaries

A piecewise-defined function is specified by different formulas on different parts of its domain. The value where the formula changes is called a breakpoint. For example, a boundary may be given by one expression for \(x\leq 1\) and another for \(x>1\); its breakpoint is \(x=1\).

For a region described using vertical slices, let \(U(x)\) be the upper boundary and \(L(x)\) the lower boundary on \([a,b]\). Where \(U(x)\geq L(x)\), the height of a slice is \(U(x)-L(x)\). If the formulas change at breakpoints, the area must be written as a sum of integrals over intervals on which the formulas are fixed.

$$ A=\sum_{j=1}^{n}\int_{x_{j-1}}^{x_j}\bigl(U(x)-L(x)\bigr)\,dx, \qquad a=x_0<x_1<\cdots<x_n=b. $$

The partition points include every breakpoint of either boundary that lies in the interval. If the curves exchange order inside a partition interval, that point must also be included so the upper-minus-lower description remains valid. The earlier material on boundary switches treats that situation; here we focus on changes in the boundary formulas.

For horizontal slices, let \(R(y)\) be the right boundary and \(L(y)\) the left boundary. Provided \(R(y)\geq L(y)\), each slice has width \(R(y)-L(y)\), so the area is found by integrating that width with respect to \(y\), again splitting at all formula breakpoints.

Why Splitting the Integral Works

On any interval where the boundary formulas are fixed and correctly ordered, the area of a thin vertical slice is its height multiplied by its small width. The height is \(U(x)-L(x)\). Adding the areas of the slices and taking the limit gives the definite integral of that height.

When a boundary changes formula, the height is described by a different expression on the other side of the breakpoint. The total area is the sum of the areas on the two sides. The additivity property of definite integrals says that splitting an interval at a point does not change the total integral; it writes that total as the sum of the integrals over the pieces. Applying this property at each breakpoint gives the formula above.

Justification: On each subinterval, the area is the integral of the distance between the two boundaries. The finitely many subintervals cover the full interval without overlapping interiors, so adding their areas gives the area of the whole region.

A breakpoint is needed even if the two formulas happen to give the same value there. The formula may still differ on either side, so a single expression for the slice height may not apply across it. Conversely, the value assigned to a boundary at one isolated breakpoint does not affect the definite integral or the area; changing a function at finitely many points leaves its integral unchanged.

Worked Example: A Piecewise Upper Boundary

Worked Example: A Piecewise Upper Boundary

On \(0\leq x\leq 2\), a region has lower boundary \(L(x)=x+1\) and upper boundary

$$ U(x)= \begin{cases} 2x+3, & 0\leq x\leq 1,\\ 6-x, & 1<x\leq 2. \end{cases} $$

Find its area and check the boundary order.

The upper boundary changes formula at \(x=1\), so split the interval into \([0,1]\) and \([1,2]\). On the first interval, the vertical distance is

$$ U(x)-L(x)=(2x+3)-(x+1)=x+2. $$

This is positive on \([0,1]\), since \(x+2\geq 2\). On the second interval, the distance is

$$ U(x)-L(x)=(6-x)-(x+1)=5-2x. $$

For \(1\leq x\leq 2\), this is at least \(5-2(2)=1\), so the stated upper boundary remains above the lower boundary. Now integrate each expression:

$$ \begin{aligned} A &=\int_0^1(x+2)\,dx+\int_1^2(5-2x)\,dx\\ &=\left[\frac{x^2}{2}+2x\right]_0^1 +\left[5x-x^2\right]_1^2\\ &=\left(\frac12+2\right)-0 +\bigl((10-4)-(5-1)\bigr)\\ &=\frac52+(6-4)\\ &=\frac92. \end{aligned} $$

Check: the two pieces contribute \(\frac52\) and \(2\), whose sum is \(\frac92\). The second integral uses the upper formula \(6-x\) and the lower formula \(x+1\), so its integrand is their difference, \(5-2x\), as required.

When Both Boundaries Have Breakpoints

If both boundaries are piecewise-defined, make one partition from the breakpoints of both functions. For instance, if the upper boundary changes at \(x=1\) and the lower boundary changes at \(x=1.5\), the intervals are \([0,1]\), \([1,1.5]\), and \([1.5,2]\). On each interval, select the applicable formula from each boundary, subtract lower from upper, and check the result is nonnegative.

1
List the breakpoints.
Include every formula change of either boundary that lies within the region’s interval.
2
Form the partition.
Put the breakpoints in order between the endpoints of the full interval.
3
Write each slice distance.
Use the appropriate upper and lower formulas on each subinterval.
4
Check order and add.
Confirm the distances are nonnegative, integrate on each interval, and add the results.

Worked Example: Breakpoints in Both Boundaries

Worked Example: Breakpoints in Both Boundaries

For \(0\leq x\leq 2\), let the upper boundary be

$$ U(x)= \begin{cases} 3, & 0\leq x\leq 1,\\ x+2, & 1<x\leq 2, \end{cases} $$

and the lower boundary be

$$ L(x)= \begin{cases} x, & 0\leq x\leq \frac32,\\ \frac{x}{2}+\frac34, & \frac32<x\leq 2. \end{cases} $$

The breakpoints are \(1\) and \(\frac32\). Use the intervals \([0,1]\), \([1,\frac32]\), and \([\frac32,2]\). On the first interval, the height is \(3-x\), which is at least \(2\). On the second, it is \((x+2)-x=2\). On the third, it is

$$ (x+2)-\left(\frac{x}{2}+\frac34\right) =\frac{x}{2}+\frac54, $$

which is positive throughout \([\frac32,2]\). Thus the proposed upper and lower boundaries are in the correct order on all three pieces. The area is

$$ \begin{aligned} A &=\int_0^1(3-x)\,dx +\int_1^{3/2}2\,dx +\int_{3/2}^2\left(\frac{x}{2}+\frac54\right)\,dx\\ &=\left[3x-\frac{x^2}{2}\right]_0^1 +\left[2x\right]_1^{3/2} +\left[\frac{x^2}{4}+\frac{5x}{4}\right]_{3/2}^2\\ &=\left(3-\frac12\right)+(3-2) +\left(\left(1+\frac52\right) -\left(\frac{9}{16}+\frac{15}{8}\right)\right)\\ &=\frac52+1+\left(\frac{56}{16}-\frac{39}{16}\right)\\ &=\frac52+1+\frac{17}{16}\\ &=\frac{73}{16}. \end{aligned} $$

Check: the three nonnegative contributions are \(\frac52\), \(1\), and \(\frac{17}{16}\). Their sum is \(\frac{40}{16}+\frac{16}{16}+\frac{17}{16}=\frac{73}{16}\), matching the calculated area.

Piecewise Left and Right Boundaries

The same partitioning principle applies to horizontal slices. The difference is that each slice’s length is measured horizontally, from the left boundary to the right boundary. If either of those boundaries changes formula as \(y\) varies, split the \(y\)-interval at its breakpoint.

Worked Example: Piecewise Left and Right Boundaries

Worked Example: Piecewise Left and Right Boundaries

For \(0\leq y\leq 2\), suppose the left boundary is

$$ L(y)= \begin{cases} y, & 0\leq y\leq 1,\\ \frac{y+1}{2}, & 1<y\leq 2, \end{cases} $$

and the right boundary is

$$ R(y)= \begin{cases} 3, & 0\leq y\leq \frac32,\\ 4-\frac{2y}{3}, & \frac32<y\leq 2. \end{cases} $$

The breakpoints are \(y=1\) and \(y=\frac32\). On \([0,1]\), the width is \(3-y\), which is at least \(2\). On \([1,\frac32]\), the width is \(3-\frac{y+1}{2}=\frac52-\frac y2\), which is positive. On \([\frac32,2]\), it is

$$ \left(4-\frac{2y}{3}\right)-\frac{y+1}{2} =\frac72-\frac{7y}{6}. $$

This last expression decreases over the interval, and at \(y=2\) it equals \(\frac72-\frac73=\frac76>0\). So the right boundary is indeed to the right of the left boundary on all three intervals. Integrate the widths:

$$ \begin{aligned} A &=\int_0^1(3-y)\,dy +\int_1^{3/2}\left(\frac52-\frac y2\right)\,dy +\int_{3/2}^2\left(\frac72-\frac{7y}{6}\right)\,dy\\ &=\left[3y-\frac{y^2}{2}\right]_0^1 +\left[\frac{5y}{2}-\frac{y^2}{4}\right]_1^{3/2} +\left[\frac{7y}{2}-\frac{7y^2}{12}\right]_{3/2}^2\\ &=\frac52+\frac{15}{16}+\frac{35}{48}\\ &=\frac{120}{48}+\frac{45}{48}+\frac{35}{48}\\ &=\frac{200}{48}\\ &=\frac{25}{6}. \end{aligned} $$

Check: each width is positive, and the three areas add to \(\frac{25}{6}\). The horizontal-slice setup uses right minus left, as required.

A Common Mistake: Extending One Formula Too Far

It is tempting to write one integral over the entire region using the first formula encountered. In the first worked example, that would incorrectly use the first upper-boundary formula \(2x+3\) all the way from \(0\) to \(2\). The resulting calculation would be

$$ \int_0^2\bigl((2x+3)-(x+1)\bigr)\,dx =\int_0^2(x+2)\,dx =\left[\frac{x^2}{2}+2x\right]_0^2 =6. $$

This is wrong because \(2x+3\) is not the upper-boundary formula on \(1<x\leq 2\). The correction is to split at \(x=1\), use the correct formula on each interval, and obtain \(\frac92\), as calculated above. A related error is to subtract in the wrong order: for vertical slices use upper minus lower, and for horizontal slices use right minus left. Always check that the resulting distance is nonnegative before integrating.

Why the Breakpoints Matter

Splitting at formula breakpoints is more than a bookkeeping step. It ensures that every integral represents the actual slice dimensions of the region. It also gives a practical check: on each interval, the selected formulas should describe the correct boundaries and yield a nonnegative height or width. If that check fails, revisit the formula choices, the interval, or the boundary order before calculating.

Takeaway: Gather the breakpoints from both boundaries, split the full interval at each one, use the appropriate formulas on each piece, verify the boundary order, and add the resulting integrals.

Check Your Understanding

For each question, identify the relevant partition and explain how you would check the boundary order.

  1. If the upper boundary changes formula at \(x=2\) and the lower boundary at \(x=3\), how should the interval \([1,4]\) be partitioned?
  2. For a vertical-slice area, which boundary is subtracted from which? What inequality should the resulting height satisfy?
  3. If two boundaries have a breakpoint at the same coordinate, should that coordinate appear once or twice in the partition?
  4. For horizontal slices, what is the expression for slice width, and what inequality should it satisfy?
  5. Why can a boundary formula not generally be used across an interval on which its definition changes?

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